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Katena32 [7]
3 years ago
10

Susan and Tom want to know the combined amount of money they earned last week. They both earned $9 per hour, but Susan also earn

ed a $45 bonus. Susan added her earnings to Tom's earnings to write the expression (9S + 45) + 9T, where S is the number of hours she worked, and Tis the number of hours Tom worked. Tom added their combined hourly wages to Susan's bonus to write the expression 9(S + T) + 45. Which of the following statements is true?

Mathematics
1 answer:
Anna007 [38]3 years ago
3 0

Answer:

A - both correct

Step-by-ste

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It would be B. Diagonal
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A school bus has 25 seats, with 5 rows of 5 seats. 15 students from the first grade and 5 students from the second grade travel
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Number of ways the second grade students can sit on the first row is 5! = 5P5
Number of ways 15 first grade students can sit in the remaining 20 seats is 20P15

Therefore, required number of ways is 5P5 * 20P15.  (option B)
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Given a conditional statement p → q, find the inverse of its inverse, the inverse of its converse, and the inverse of its contra
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The inverse of an implication p ⇒ q is ¬p ⇒ ¬q.

The converse of p ⇒ q is q ⇒ p.

The contrapositive of p ⇒ q is ¬q ⇒ ¬p.

Then

• the inverse of the inverse of p ⇒ q is p ⇒ q

• the inverse of the the converse of p ⇒ q is ¬q ⇒ ¬p

• the inverse of the contrapositive of p ⇒ q is q ⇒ p

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What does "Uninsured Motorist Bodily Injury" typically cover
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Uninsured motorist bodily injury, or UMBI, pays for medical bills, pain and suffering, lost wages if you can't work after an accident and funeral expenses after a crash with an at-fault driver who doesn't have car insurance. It may also cover you if you're hit as a pedestrian or while riding your bike.

Step-by-step explanation:

8 0
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Past records indicate that the probability of online retail orders
Tcecarenko [31]

Answer:

a) Mean = 1.6, standard deviation = 1.21

b) 18.87% probability that zero online retail orders will turn out to be fraudulent.

c) 32.82% probability that one online retail order will turn out to be fraudulent.

d) 48.31% probability that two or more online retail orders will turn out to be fraudulent.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

The mean of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

In this problem, we have that:

p = 0.08, n = 20

a. What are the mean and standard deviation of the number of online retail orders that turn out to be fraudulent?

Mean

E(X) = np = 20*0.08 = 1.6

Standard deviation

\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{20*0.08*0.92} = 1.21

b. What is the probability that zero online retail orders will turn out to be fraudulent?

This is P(X = 0).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{20,0}.(0.08)^{0}.(0.92)^{20} = 0.1887

18.87% probability that zero online retail orders will turn out to be fraudulent.

c. What is the probability that one online retail order will turn out to be fraudulent?

This is P(X = 1).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{20,1}.(0.08)^{1}.(0.92)^{19} = 0.3282

32.82% probability that one online retail order will turn out to be fraudulent.

d. What is the probability that two or more online retail orders will turn out to be fraudulent?

Either one or less is fraudulent, or two or more are. The sum of the probabilities of these events is decimal 1. So

P(X \leq 1) + P(X \geq 2) = 1

We want P(X \geq 2)

So

P(X \geq 2) = 1 - P(X \leq 1)

In which

P(X \leq 1) = P(X = 0) + P(X = 1)

From itens b and c

P(X \leq 1) = 0.1887 + 0.3282 = 0.5169

P(X \geq 2) = 1 - P(X \leq 1) = 1 - 0.5169 = 0.4831

48.31% probability that two or more online retail orders will turn out to be fraudulent.

4 0
4 years ago
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